Solve PDE u_t=u_xx-u_x with Separation of Variables

[Uku]

Hi! Merry christmas!

Homework Statement

$$u_{t}=u_{xx}-u_{x}$$

Can I solve it with separation of variables?

The Attempt at a Solution

$$u=XT$$
$$XT^{'}=T(X^{''}-X^{'})$$
After rearranging
$$\frac{T^{'}}{T}=-\lambda^{2}$$ 1)
$$\frac{X^{''}}{X} - frac{X^{'}}{X}=-\lambda^{2}$$
The solution to 1) is simple
$$T=Acos(\lambda t)+Bsin(\lambda t)$$
Now, for the X, I write out a characteristic equation and get that
$$X=Ce^{\frac{1}{2}+\sqrt{\frac{1}{4} - \lambda^{2}}x}+De^{\frac{1}{2}-\sqrt{\frac{1}{4} - \lambda^{2}}x}$$
Since u=XT, the solution would be:
$$XT=(Acos(\lambda t)+Bsin(\lambda t))(Ce^{\frac{1}{2}+\sqrt{\frac{1}{4} - \lambda^{2}}x}+De^{\frac{1}{2} - \sqrt{\frac{1}{4} - \lambda^{2}}x})$$

Now, have I done it wrong?
(I know that I have not checked, I will do it now)

Thanks,
Uku

 

[HallsofIvy]

Hi! Merry christmas!

Homework Statement

$$u_{t}=u_{xx}-u_{x}$$

Can I solve it with separation of variables?

The Attempt at a Solution

$$u=XT$$
$$XT^{'}=T(X^{''}-X^{'})$$
After rearranging
$$\frac{T^{'}}{T}=-\lambda^{2}$$ 1)
$$\frac{X^{''}}{X} - frac{X^{'}}{X}=-\lambda^{2}$$

Do you know that this constant is negative? That usually is determined by boundary conditions and you don't give any here.

[tex]The solution to 1) is simple<br /> $$T=Acos(\lambda t)+Bsin(\lambda t)$$<br /> Now, for the X, I write out a characteristic equation and get that<br /> $$X=Ce^{\frac{1}{2}+\sqrt{\frac{1}{4}-}\lambda^{2}}+De^{\frac{1}{2}-\sqrt{\frac{1}{4} - \lambda^{2}}}$$[/tex]

[tex] Only if $\lambda^2< 1/4$, otherwise you will have trig functions. Also you have forgotten the variable, x!<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Since u=XT, the solution would be:<br /> $$XT=(Acos(\lambda t)+Bsin(\lambda t))(Ce^{\frac{1}{2}+\sqrt{\frac{1}{4}-}\lambda^{2}}+De^{\frac{1}{2}-\sqrt{\frac{1}{4} - \lambda^{2}}})$$<br /> <br /> Now, have I done it wrong? <br /> (I know that I have not checked, I will do it now)<br /> <br /> Thanks,<br /> Uku </div> </div> </blockquote> What boundary and initial conditions do you have? With partial differential equations, even the form of the solution will depend on them.[/tex]

 

[tiny-tim]

Hi Uku!

(have a lambda: λ )

You're getting confused between λ and λ² in both your X and T equations.

 

[Dickfore]

The equation is linear and with constant coefficients. If you present it in the form:

$$u(x, t) = e^{a x + b t} \, v(x, t)$$

you can choose a and b to cancel the first partial derivative w.r.t. x and the term with no partial derivatives:

$$v_{t} + b v = v_{x x} +2 a v_{x} + a^{2} v - v_{x} - a v$$

Choosing:

$$\left\{\begin{array}{l}<br /> 2 a - 1 = 0 \\<br /> <br /> a^2 - a = b<br /> \end{array}\right.$$

after which the equation becomes a pure parabolic equation:

$$v_{t} = v_{x x}$$

You can solve this PDE by the method of integral transforms. For example, represent the x-dependence as a Fourier transform:

$$v(x, t) = \int_{-\infty}^{\infty}{\tilde{v}(k, t) \exp(i k x) \, \frac{d k}{2 \pi}}$$

after which you get the 1st order ODE

$$\tilde{v}_{t} + k^{2} \, \tilde{v} = 0$$

which has the simple solution:

$$\tilde{v}(t, k) = \tilde{v}(0, k) e^{-k^{2} t}$$

where the initial condition is given by:

$$\tilde{v}(0, k) = \int_{-\infty}^{\infty}{v(0, \xi) \exp(-i k \xi) d\xi}, \; v(0, x) = e^{-a x} u(0, x)$$

After you combine everything together and do the integral over k which is gaussian, you will get the general solution as an integral over the initial value.

 

[Uku]

Hi!

I messed up on the lamdas, yes. On $$T$$ it should be just $$\lambda$$.
My initial conditions are:
$$-\infty < x < \infty$$
$$0 < t < \infty$$
$$u(x,0)=f(x)$$

How do these tell me wether lamda is negative or not? I got a book, but I'm not getting to the point from it.

EDIT: I forgot the x from the exponential in the initial post, I added it.

 

[tiny-tim]

Hi Uku!

(what happened to that λ I gave you? )

How do these tell me wether lamda is negative or not?

λ is a constant of integration.

It can be anything … you find it from your initial (or boundary) conditions …

what are they?​

 

[Uku]

Tim, thanks for the lamda? :)

I have an initial condition
$$u(x,0)=f(x)$$
telling me that at t=0 the temperature (the equation looks as a heat conduction eq for a bar of something) depends on position only.

Then I have boundary conditions for the variables
$$- \infty < x < \infty$$
telling me that the bar is infinite in both directions, and
$$0 < t < \infty$$
telling me that the time changes from 0 to inf.

As I understand the lamda comes from assuming that once the variables are separated, each side depends on only X (or Y). For the equality to hold the constant needs to be the same for both sides, hence the X'/X=c Y'/Y=c part.

If I try to link a boundary to c, I can think of, in
X'-Xc=0
that c scales the position X, so in that case c also must subdue to
-infinity < c < infinity?

I don't yet see a clear link with the initial (bound.) conditions.

Uku

 

[tiny-tim]

Hi Uku!

What is f(x)?

On its own, u(x,0)=f(x) is only stating the obvious … "u(x,0) does not depend on t".

And (-∞.∞) isn't a boundary condition, it's only a boundary.

 

[Hellohi]

dont mind me, keep explaining.:zzz: